What is the solution for 2y^{''}+35y=0 ?

The solution for 2y^{''}+35y=0 is y=c_{1}cos(sqrt(35/2)t)+c_{2}sin(sqrt(35/2)t)

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2y^{''}+35y=0

2 y^{^{''}} + 35 y = 0

y = c_{1} cos (\frac{35}{2} t) + c_{2} sin (\frac{35}{2} t)

Solution steps

2 y^{''} (t) + 35 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{35}{2} t) + c_{2} sin (\frac{35}{2} t)

y = c_{1} cos (\frac{35}{2} t) + c_{2} sin (\frac{35}{2} t)

x \to 2 lim (\frac{x ^{2}}{x ^{2} + 1})

\int e^{- in x} d x

\int (sin (x) - cos (x)) d x

\int x^{2} cos (n) x d x

x \to 6 lim (- 2 x - 5)

What is the solution for 2y^{''}+35y=0 ?
The solution for 2y^{''}+35y=0 is y=c_{1}cos(sqrt(35/2)t)+c_{2}sin(sqrt(35/2)t)